DMSO is an ambidentate ligand that can bind in a κO fashion via the oxygen lone pair and in a κS fashion via the sulfur lone pair.
I know that DMSO can behave as a π-acceptor when it bonds a metal at the sulfur, but what is the symmetry of the $\ce{O\bond{->}M}$ MO when DMSO binds at the oxygen? I have some diagrams of DMSO's HOMO and LUMO MOs, and it seems clear that DMSO-κS is a π-acceptor due to the LUMO's potential to overlap with a $\mathrm{t_{2g}}$-type d orbital, but the HOMO looks like it would do a π interaction. This confuses me because I know from IR that the $\ce{S-O}$ bond is weakened when DMSO binds at the oxygen, and there are some lower energy occupied MOs that would have σ symmetry overlapping with $\mathrm{d}_{z^2}$, for instance.
Is DMSO-κO a σ-donor, a π-donor, or both?
(keep in mind that there are some more σ-donor-looking MOs at energy levels below the HOMO)
Also; The reference below (1) says that DMSO's tendency to bind as DMSO-κS is dependent on the metal having a high enough "electron charge density" to have a sufficient back donation contribution to the $\ce{M\bond{->}S}$ π bond. What does this mean? The paper also says that ruthenium and rhodium prefers to bind as DMSO-κS. Based on Pearson's HSAB concept, I would expect that ruthenium would prefer DMSO-κS because it is a soft acid as a result of its larger second-row ionic radius giving it lower charge density (and S is softer than O), but the paper seems to be saying the opposite. What do they mean by "electron charge density"?
(1) Panina, N. S; Calligaris, M. Inorg. Chim. Acta 2002, 334, 165-171.